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0 votes
58.6k views
in Mathematics by (67.9k points)

Locus of the midpoint of any focal chord of y2 = 4ax is 

(a)  y2 = a(x – 2a) 

(b)  y2 = 2a(x – 2a)

(c)  y2 = 2a(x – a)

(d)  none of these

1 Answer

+1 vote
by (63.3k points)
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Best answer

Correct option (c) y2 = 2a(x – a)

Explanation:

Let the midpoint be P(h, k).

Equation of this chord is

T = S1 . i.e., yk – 2a(x + h) = k2 – 4ah.

It must pass through (a, 0)

⇒ 2a(a + h) = k2 – 4ah.

Thus required locus is y2 = 2ax – 2a2 .

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